NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR …...NSC/NSS – Marking Guidelines/Nasienriglyne...

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These marking guidelines consist of 18 pages.

Hierdie nasienriglyne bestaan uit 18 bladsye.

MATHEMATICS P1/WISKUNDE V1

NOVEMBER 2019

MARKING GUIDELINES/NASIENRIGLYNE

GRADE 12/GRAAD 12

NATIONAL

SENIOR CERTIFICATE/

NASIONALE SENIOR

SERTIFIKAAT

Downloaded from Stanmorephysics.com

Mathematics P1/Wiskunde V1 2 DBE/November 2019

NSC/NSS – Marking Guidelines/Nasienriglyne


NOTE:

If a candidate answers a question TWICE, only mark the FIRST attempt.

Consistent Accuracy applies in all aspects of the marking memorandum.

LET WEL:

Indien 'n kandidaat ’n vraag TWEE keer beantwoord, merk slegs die EERSTE poging.

Volgehoue akkuraatheid is op ALLE aspekte van die nasienriglyne van toepassing.

QUESTION/VRAAG 1

1.1.1 0652 xx

0)1)(6( xx

1or6 xx

factors

6x 1x (3)

1.1.2 0534 2 xx

a

acbbx

2

42

)4(2

)5)(4(4)3(3 2 x

8

893x

8,0or55,1 xx

substitution into the

correct formula

55,1x 8,0x (3)

1.1.3 014 2 x

0)12)(12( xx

2

1

2

1

x

factors

method

answer (3)

1.1.4 xxx

3232

22

2

32

32

xx

xx

4

16

322

2

2

x

x

x

4x

232 x

squaring both sides

162 x

4x (selection) (4)

2

1

2

1

+ + –




1.2 12 xy

)1(..........12 xy

)2........(314 xxy

Sub (1) into (2)

xxx 314)12(

0314122 xxx

014152 xx

014152 xx

0)1)(14( xx

1or14 xx

11or2 yy

OR/OF

12 xy

)1(..........12 yx

)2.........(314 xxy

Sub (1) into (2)

)12(314)12( yyy

03361412 2 yyy

02292 yy

02292 yy

0)11)(2( yy

11or2 yy

1or14 xx

y subject of the formula

substitution

simplification

both values of x

both values of y (5)

OR/OF

x subject of the formula

substitution

simplification

both values of y

both values of x (5)

1.3

3 6 9 12 15 18 21 24 27 30

3 3 23 3 3 23 3 3 33 3

14k

identifying multiples of 3

ten multiples of 3

powers of 3

answer (4)

[22]





QUESTION/VRAAG 2

2.1.1 209 ; 186 209 186 (2)

2.1.2

35434

321)34(131)1(31

32131322

222

2729311

234;261;290;321

cb

cba

cbabaa

diffnd

diffst

354342 nnTn

2nd

diff = 2

a = 1

b = –34

c = 354

(4)

2.1.3 74354342 nn

0280342 nn

02014 nn

20or14 nn

equating nT to 74

standard form

14 20 (4)

2.1.4

17

342

0342

0)(/

n

n

n

nf

Term 17 will have the smallest value

OR/OF

17

2

34

2

n

n

a

bn


OR/OF

172

2014

n


0342 n

answer )2(

OR/OF

substitution

answer )2(

OR/OF

substitution

answer )2(




2.2.1

251

24991

2

11

2

11

8

5

1

1

212

1

8

5

21

21

,

...,

S

r

)r(aS

n;r;a

n

n

r


correct formula

answer (3)

2.2.2

1010

1111

101101

222

1

2

1

1024

1

2

1

8192

5

2

1

8

5

8192

5

8192

5

101

101

1

1

1

nn

nn

nn

or

ar

T

n

n

n

n

n

n

OR/OF

8 ; 16 ; 32 ; … ; 8192

10

11

101

22

10242

81922.8

101

1

1

n

n

n

n

n

n

substitution into the correct

formula

method /same base or log

calculating n

answer

(4)

OR/OF

substitution into the correct

formula

method

calculating n

answer (4)

[19]





QUESTION/VRAAG 3

3.1

10

3

10

3 1

1

2

1

y y yy

=

9

1

8

1....

3

1

2

1

8

1....

3

1

2

1

1

1

= 1–9

1

= 9

8

8

1....

3

1

2

1

1

1

9

1

8

1....

3

1

2

1

answer (3)

3.2

3

8....

3

2

9

4

9

2

3

24....

3

21

3

2

3

2

3

2

3

1

9

2a

9

2

9

4

3

2 and d

dnan

Sn )1(22

OR lan

Sn 2

9

2)112(

9

22

2

1212S

3

8

9

2

2

1212S

2m3

52 2m

3

52

for both sides = 2 × 2m67,343

104

3

52

OR/OF

2m 67,34

2)121110987654321(9

2

OR/OF

3

812

9

21 T

9

21

9

2l

2

12

m 67,34

9

2

3

8

2

1222

S

a

d


correct formula

answer

answer for both sides

(6)

OR/OF

a

(1 + …. + 12)

2

answer (6)

OR/OF

a

3

81 T

9

2l

substitution into correct

formula

answer (6)

[9]




QUESTION/VRAAG 4

4.1 1p p = –1 (1)

4.2

1

x

ay

103

a

3a

32 bxxy

3110 2 b 2b

coordinates D(0 ; –3)

substitute (0 ; –3)

substitute (1 ; 0)

(3)

4.3 322 xxy

a

bx

2 :symofaxis

)1(2

2x

1x

43)1(2)1( 2 y

)4;1(C

OR/OF

0dx

dy

022 x

1x

43)1(2)1( 2 y

)4;1(C

substitution

1x

substitution

y = –4 (4)

OR/OF

derivative

1x

substitution

y = –4 (4)

4.4 4or);4[ yy 4

answer (2)

4.5 145tan m

cmxy

c )1)(1(4

3c

3 xy

gradient

subs m and )4;1(

equation (3)

4.6 No, the line passes through C and D

OR/OF

No, a tangent through turning point C will have a

gradient of 0

No

reason (2)

OR/OF

No

reason (2)





4.7 mxfxmf f is reflected in the y-axis and translated 1 unit to

the left and 4 units upwards.

Therefore: 1m

4q

OR/OF

Substitute 4and0 qx for one x- intercept

4

1

)1(0

120

43)0(2)0()0(

3)(2)()(

2

2

2

2

q

m

m

mm

mmh

qxmxmxh

value of m

value of q (4)

OR/OF

value of m

value of q (4)

[19]




QUESTION/VRAAG 5

5.1 xkxf )(

416 k

2k

substitution

(4 ; 16)

answer (2)

5.2

xy

xf

yf

y

x

2

1

log

2 :

2 :

yx 2

xy 2log

(2)

5.3

asymptote

shape

for any two

valid

points

eg.(16 ; 4) or

(2 ; 1) or (4 ; 2)

or (1 ; 0)

(4)

5.4.1 );1( x or 1x 1

answer

(2)

5.4.2 2

10 x or

2

1;0x

2

1

answer

(2)

x

y

(16 ; 4)

(1 ; 0) .





5.5

2 N/A

22or4

12

042or012.4

0)42)(12.4(

042.152.4

21542.4

24

1512

4

15

2

12

4

1522

2

2

2

2

x

xx

xx

xx

xx

xx

xx

x

x

xx

OR/OF

2 N/A

22or4

12

4or4

1

0)4)(1.4(

04.15.4

154.4

4

151

2Let

4

15

2

12

4

1522

2

2

2

2

x

kk

kk

kk

kk

kk

k

xx

x

x

x

xx

4

1522 xx

standard form

factors

answer

(4)

OR/OF

4

1522 xx

standard form

factors

answer

(4)

[16]




QUESTION/VRAAG 6

6.1 )1(:Kuda inPA

)4083,01(0005

00,6606R

R266,40660,00R6 :AnswerFinal

40,9266R

OR/OF

04,1)1(:Kuda inPA

04,1)4083,01(0005

40,9266R

niPA )1(:Thabo

412

12

081,010005

71,9056R

Kuda will have a better investment


correct formula

final answer

OR/OF


correct formula

final answer


correct formula

answer

conclusion (5)

6.2.1

i

ixP

n ])1(1[

12

1,0

12

1,0110006

000525

n

48

13log

12

1,01log

12

1,011

48

35

n

n

n

12

1,01log

48

13log

40,157n

n = 158 payments

OR/OF

12

1,0


correct formula

simplification

use of logs

answer (5)

OR/OF





i

ixP

n ])1(1[

12

1,0

12

1,0110006

000525

12

n

48

13log

12

1,01log12

12

1,011

48

3512

n

n

n12

12

1,01log

48

13log

n12

1

12

1,01log

48

13log

11686841,13n

Number of payments = 13,1168684112=157,40

n = 158 payments

12

1,0


correct formula

simplification

use of logs

answer

(5)

6.2.2 Difference: R6 000 – R5 066,36 = R933,64

i

ixF

n ]1)1[(

51,503162

12

1,0

112

1,0164,933

108

R

F

OR/OF

R933,64

n = 108


correct formula

answer (4)

OR/OF




i

ixF

n ]1)1[(

12

1,0

112

1,010006

108

F

28,3220441R

12

1,0

112

1,0136,0665

108

F

.....77,818188RF

Amount available for withdrawal

= R1 044 322,28 – R 881 818,77

= R162 503,51

OR/OF

Outstanding balance with monthly repayment of

R5 066,35

12

1,0

112

1,0136,0665

12

1,01000525

108

108

= R404 666, 23

Outstanding balance with monthly repayment of

R6 000

12

1,0

112

1,010006

12

1,01000525

108

108

= R242 162,72

Amount available for withdrawal

R404 666,23 – R242 162,72 = R162 512,18

n = 108

substitution

into correct formula

substitution into

correct formula

final answer

(4)

OR/OF

n = 108


correct formula


correct formula

final answer

(4)

[14]





QUESTION/VRAAG 7

7.1 xxf 74)(

h

xfhxfxf

h

)()(lim)(

0

/

h

xhx

h

)74()(74lim

0

h

h

h

)7(lim

0

= –7

)(74 hx

substitution

simplification

answer (4)

7.2 384 xxy

23

84 xx

2

1

7

2

332 xx

dx

dy

23

x

732x

21

2

3x (3)

7.3.1 aaxy 2

02 axdx

dy

axdx

dy2

ax2 (1)

7.3.2 aaxy 2

12 xda

dy

answer (2)




7.4 Substitute (2 ; b) in

xxy

12

2

122b

8b

dx

dym tangent

2

121

xdx

dy

22

121

2tangentm

2

1perp m

Equation of perpendicular line:

11 xxmyy OR cmxy

22

18 xy c 2

2

18

72

1 xy 7c

72

1 xy

value of b

2

121

xdx

dy

gradient of perpendicular

line

equation (4)

[14]

QUESTION/VRAAG 8

8.1 36cm answer (1)

8.2 )632(6 2 ttt have no real roots

Insect reaches the floor only once.

only once (3)

8.3 3624152)( 23 tttth

24306)( 2 ttth

024306 2 tt

0452 tt

0)1)(4( tt

1or4 tt

Only 4t because maximum value required

cmh 5236)4(24)4(15)4(2 23

expansion

24306 2 tt = 0

both values

answer (4)

[8]





QUESTION/VRAAG 9

9.1

3or 0

0)3(3

093

93

9)(

2

23

23

2/

x x

xx

xx

xx

xxf

2/ 9)( xxf

x = 0

x = 3 (3)

9.2.1 For f and /f answer (1)

9.2.2 The point (0 ; 0) is :

A point of inflection of f

A turning point of /f

f : inflection

point

point turning:/f (2)

9.3 xxf 18)(//

)1()1(Distance /// ff

2)1(9)1(18

9

xxf 18)(//

substitution

answer (3)

9.4

03but

0)3(3

093

2

2

23

x

xx

xx

0,3

03

xx

x

093 23 xx

factors

3x

0x (4)

[13]

QUESTION/VRAAG 10

10.1 P(same day) =16

4 or

4

1 or 0,25 or 25%

4 numerator

16 denominator (2)

10.2 P(2 consecutive days) = 8

3

16

23

3 2

answer (3)

[5]

0 3




QUESTION/VRAAG 11

11.1.1 )()( BPAP independent events

25,040,0 1,0

0,1

0,15 and 0,3

0,45 (3)

11.1.2 BnotandABnotA)BnotorA( PPPP

3,075,04,0

85,0

OR/OF

P(A or not B) = 1 – P(only B)

= 1 – 0,15

= 0,85

OR/OF

From Venn diagram:

0,3 + 0,1 + 0,45 = 0,85

substitution

answer

(2)

OR/OF

1 – 0,15

answer

(2)

OR/OF

substitution

answer (2)

11.2 (5 × 1 × 5) + (5 × 1 × 6 )+ (5 × 1 × 6) + (5 × 1 × 5) =110

110 × 5 = 550 > 500

Not possible, because not enough space

OR/OF

(5 × 2 × 5) + (5 × 2 × 6) = 110

110 × 5 = 550 > 500

Not possible because not enough space

OR/OF

5 × 1 × 5

5 × 1 × 6

5 × 1 × 6

5 × 1 × 5

110

conclusion (6)

OR/OF

5 × 2 × 5

5 × 2 × 6

110

conclusion (6)

OR/OF

A B

0,3 0,1 0,15

0,45




Copyright reserved/Kopiereg voorbehou

11010120

1025

120645

110 × 5 = 550 > 500

Not possible because not enough space

120645

1025

10120

110

conclusion (6)

[11]

TOTAL/TOTAAL: 150

NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR …...NSC/NSS – Marking Guidelines/Nasienriglyne...

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